\(\lim\limits_{x\rightarrow-1}\frac{\sqrt{x^2+3}-2}{x+1}=\lim\limits_{x\rightarrow-1}\frac{\left(\sqrt{x^2+3}-2\right)\left(\sqrt{x^2+3}+2\right)}{\left(x+1\right)\left(\sqrt{x^2+3}+2\right)}\)
\(=\lim\limits_{x\rightarrow-1}\frac{x^2-1}{\left(x+1\right)\left(\sqrt{x^2+3}+2\right)}=\lim\limits_{x\rightarrow-1}\frac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(\sqrt{x^2+3}+2\right)}\)
\(=\lim\limits_{x\rightarrow-1}\frac{x-1}{\sqrt{x^2+3}+2}=\frac{-2}{4}=-\frac{1}{2}\)
Câu 2 đề là \(\lim\limits_{x\rightarrow-3^+}\frac{1-x}{3+x^2}\) hay \(\lim\limits_{x\rightarrow-3^+}\frac{1-x}{\left(3+x\right)^2}\)
Đoán là bạn gõ nhầm kí tự, cái sau thì hợp lý hơn
\(\lim\limits_{x\rightarrow-3^+}\frac{1-x}{\left(x+3\right)^2}=\frac{1-\left(-3\right)}{0}=\frac{4}{0}=+\infty\)